If 30 cm³ of oxygen diffuses through a porous plug in 7 seconds, how long will it take 60 cm³ of chlorine to diffuse through the same plug?

Steps to Solve

1. State Graham's Law of Effusion

Graham's law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. The formula is:

$$ \frac{r_1}{r_2} = \frac{\sqrt{M_2}}{\sqrt{M_1}} $$

where $r_1$ and $r_2$ are the rates of diffusion for gas 1 and gas 2, respectively, and $M_1$ and $M_2$ are their molar masses.

2. Identify the gases and their molar masses

In this problem, we have:

• Oxygen ($O_2$) with a molar mass of about 32 g/mol ($M_1 = 32$).
• Chlorine ($Cl_2$) with a molar mass of about 71 g/mol ($M_2 = 71$).

3. Set up the ratio of diffusion rates

Since we know the time taken for oxygen to diffuse, we can set $\frac{r_{O_2}}{r_{Cl_2}}$ as:

$$ \frac{r_{O_2}}{r_{Cl_2}} = \frac{\sqrt{M_{Cl_2}}}{\sqrt{M_{O_2}}} $$

4. Substitute known values into the equation

We can plug in the values of the molar masses:

$$ \frac{r_{O_2}}{r_{Cl_2}} = \frac{\sqrt{71}}{\sqrt{32}} $$

5. Calculate the rates of diffusion

Now we can calculate the value of the expression:

$$ \frac{r_{O_2}}{r_{Cl_2}} = \frac{\sqrt{71}}{4\sqrt{2}} \approx 1.24 $$

6. Relate the volumes of the gases

Given that $30 \text{ cm}^3$ of oxygen takes $7$ seconds, we can express the rate of diffusion:

$$ r_{O_2} = \frac{30 \text{ cm}^3}{7 \text{ s}} $$

And we want to find the time it takes for $60 \text{ cm}^3$ of chlorine:

$$ t_{Cl_2} = \frac{60 \text{ cm}^3}{r_{Cl_2}} $$

7. Relate time and rates

Using the relationship of rates, we find:

$$ t_{Cl_2} = t_{O_2} \times \frac{V_{Cl_2}}{V_{O_2}} \times \frac{r_{O_2}}{r_{Cl_2}} $$

Substituting the values gives:

$$ t_{Cl_2} = 7 \text{ s} \times \frac{60}{30} \times \frac{1}{1.24} $$

8. Calculate the final time

Now calculate $t_{Cl_2}$:

$$ t_{Cl_2} = 7 \text{ s} \times 2 \times \frac{1}{1.24} \approx 11.29 \text{ s} $$